advanced chemistry practice problems - mrs....

Advanced Chemistry Practice Problems

Finding pH

1. Question: Determine the pH for each of the given solutions. a. 0.150 M HNO3 b. 0.150 M CH3COOH, Ka = 1.8 × 10-5 c. 0.150 M CHOOH, Ka = 3.5 × 10-4

Answer: The method to determine the pH of a solution will depend on whether the acid is strong or weak.

a. 0.150 M HNO3 Nitric acid is a strong, monoprotic acid. Therefore, [HNO3] = [H+] = 0.150 M

0.824 pHlog[0.150] pH

]log[HpH

� � �

b. 0.150 M CH3COOH, Ka = 1.8 × 10-5

For a weak acid, such as acetic acid, an equilibrium equation and ICE table must be used to determine the pH. Weak acids do not completely dissociate so we cannot assume that [H+] = [CH3COOH]. The Ka value (given in the problem) is 1.8 × 10-5.

CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)

INITIAL: The initial concentration of CH3COOH is known to be 0.150 M. Since the problem concerns the ionization of acetic acid into its ions, the initial concentrations of the ions are zero. CHANGE: The change is represented by x and follows the stoichiometry of the balanced chemical reaction. EQUILIBRIUM: The equilibrium row is the sum of the initial and change rows for each column.


CH3COOH (M) CH3COO- (M) H+ (M) I 0.150 M 0 0 C -x +x +x E 0.150 - x x x

The law of mass action is given and the equilibrium values may be substituted into the equation to solve for x and thus the value of [H+].

> @> @

> @> @> @x0.150

xx101.8

COOHCHHCOOCH

5-

3

3a

-

K

u

��

The value of Ka is small, so the assumption that x << 0.150 (much less) can be made.

> @> @> @0.150

xx101.8 5- u

Now, the value of x can be determined.

3

62

101.6x

102.7x�

�

u

u

The value of x is less than 5% of the value from which it was subtracted (0.150) so the assumption is valid. The equilibrium value of [H+] = x, therefore [H+] = 1.6×10-3. The pH of the solution can now be calculated.

2.78pH)10log(1.6pH

]log[HpH3

u�

� �

�

The answer is reasonable because the pH of the 0.150 M HNO3 solution is lower (more acidic) than the pH of the 0.150 M CH3COOH solution. This is expected because HNO3 is a strong acid and CH3COOH is a weak acid.

c. 0.150 M CHOOH, Ka = 3.5 × 10-4 Formic acid, similar to acetic acid, is a weak acid. Therefore, an equilibrium equation and ICE table must be used to determine the pH of the solution. Weak acids do not completely ionize so we cannot assume that [H+] = [CHOOH]. The Ka value (given in the problem) is 3.5 × 10-4.


CHOOH(aq) ⇌ CHOO-(aq) + H+(aq)

INITIAL: The initial concentration of CHOOH is known to be 0.150 M. Since the problem concerns the dissociation of formic acid into its ions, the initial concentrations of the ions are zero. CHANGE: The change is represented by x and follows the stoichiometry of the balanced chemical reaction. EQUILIBRIUM: The equilibrium row is the sum of the initial and change rows for each column.

CHOOH (M) CHOO- (M) H+ (M) I 0.150 M 0 0 C -x +x +x E 0.150 - x x x


> @> @> @> @> @

> @x0.150xx10 3.5

CHOOHHCHOO

4-

a

-

K

u

��

The value of Ka is small, so the assumption that x << 0.150 can be made. > @> @> @0.150

xx103.5 4- u


3

52

102.7x

103.5x�

�

u

u

The value of x is 4.8% of the value from which it was subtracted (0.150) so the assumption is valid. The equilibrium value of [H+] = x, therefore [H+] = 7.2×10-3. The pH of the solution can now be calculated.


2.14pH)10log(7.2pH

]log[HpH3

u�

� �

�

The answer is reasonable because the pH of the 0.150 M HNO3 solution is lower (more acidic) than the pH of the 0.150 M CHOOH solution. This is expected because HNO3 is a strong acid and CHOOH is a weak acid. The pH of CHOOH is also less than CH3COOH because it is a strong acid which is evidence by the greater Ka value for CHOOH.

2. Question: A solution of 0.100 M hypochlorous acid (HOCl) has a Ka = 3.5 × 10-8,

what is the percent ionization of this solution? Answer: Solving for the percent ionization of a weak acid begins by solving for the [H+]. Once [H+] is known, the percent ionization can be calculated. HOCl is a weak acid so we cannot assume that [H+] = [HOCl]. The Ka value (given in the problem) is 3.5 × 10-8.

HOCl(aq) ⇌ OCl-(aq) + H+(aq)

INITIAL: The initial concentration of HOCl is given as 0.100 M. The initial concentrations of the ions are zero. CHANGE: The change is represented by x and follows the stoichiometry of the balanced chemical reaction. EQUILIBRIUM: The equilibrium row is the sum of the initial and change rows for each column.

HOCl (M) OCl- (M) H+ (M) I 0.100 M 0 0 C -x +x +x E 0.100 - x x x



> @> @> @> @> @

> @x0.100xx103.5

HOClHOCl

8-

a

-

K

u

��

The value of Ka is small, so the assumption that x << 0.0.100 can be made. > @> @> @0.100

xx103.5 8- u


5

92

105.9x

103.5x�

�

u

u

The value of x is less than 5% of the value from which it was subtracted (0.100) so the assumption is valid. The goal of calculating percent ionization is to determine what fraction of the acid ionized (dissociated into ions). The formula for percent ionization is

> @> @ 100

0u

�

HOClHionization %

With the initial concentration of HOCl and the calculated concentration of H+, the percent ionization can be calculated.

> @> @

% 0.059 ionization %

1000.100

105.9ionization %-5

uu

The percent ionization can serve as another way to compare the strength of acids. The higher the percent ionization, the strong the acid. HOCl is a weak acid with a very low Ka value has a very low percent ionization.


Strong and Weak Bases

1. Question: Identify each of the following as a strong or weak base. a. KOH b. NH3 c. CH3NH2 d. Ba(OH)2

Answer: Strong bases contain the hydroxide anion while most weak bases will be an ammine compound (contain an N as a central atom).

a. KOH – strong b. NH3 – weak c. CH3NH2 – weak d. Ba(OH)2 – strong

2. Question: Complete the table. Assume all solutions are at 25°C.

Solution [H+] pH [OH-] pOH HBr 0.751 HF 3.2 × 10-4

HClO4 1.04 CH2O2 5.35 × 10-10 NaOH 13.54 NH3 4.35 KOH 2.53 × 10-2

Answer: If one of the four values ([H+], [OH-], pH or pOH) is known, then the other three values can be calculated. The value of Kw at the given temperature must be known. In these examples, the temperature is 25°C and the value of Kw is 1.0×10-14.

Solution [H+] (M) pH [OH-] (M) pOH HBr 0.177 0.751 5.65 × 10-14 13.249 HF 3.2 × 10-4 3.49 3.1 × 10-11 10.51

HClO4 0.091 1.04 1.10 × 10-13 12.96 CH2O2 1.87 × 10-5 4.728 5.35 × 10-10 9.272 NaOH 2.9 × 10-14 13.54 0.46 0.34 NH3 2.2 × 10-10 9.65 4.5 × 10-5 4.35 KOH 3.95 × 10-13 12.403 2.53 × 10-2 1.597


To convert pH to [H+] [H+] = 10-pH

To convert [H+] to pH pH = -log[H+]

To convert pH to pOH pOH = 14 - pH

To convert pOH to pH pH = 14 - pOH

To convert pOH to [OH-] [OH ] = 10-pOH

To convert [OH-] to pOH pOH = -log[OH-]

To convert between [H+] and [OH-] Kw = [H+][OH-]

at 25°C 1.0 × 10-14 = [H+][OH-]

1. Question: Determine the pH for each of the given solutions. a. 0.250 M KOH b. 0.250 M Ba(OH)2 c. 0.250 M NH3, Kb = 1.8 × 10-5

Answer: To find the pH of these basic solutions, the pOH needs to be calculated first. The approach will be different for strong and weak bases.

a. KOH is a strong base. For 0.250 M KOH, the [OH-] = [KOH] therefore [OH-

] = 0.250 M

0.602pOH]log[0.250pOH

]log[OHpOH -

� �

With the value of the pOH, the pH can be determined by subtracting from 14.

13.398pH0.60214pHpOH14pH

14pOH pH

� �

�

Given that KOH is a strong base, a high pH value (close to 14) is expected.


b. For 0.250 M Ba(OH)2, the approach will be similar except for the fact that there are two hydroxide groups for every unit of Ba(OH)2 so the hydroxide concentration is twice that of the base. Therefore [OH-] = 0.500 M

0.301pOH]log[0.500pOH

]log[OHpOH -

� �

With the value of the pOH, the pH can be determined by subtracting from 14.

13.699pH0.30114pHpOH14pH

14pOH pH

� �

�

Given that the hydroxide concentration is higher for Ba(OH)2 than for KOH, the pH will be higher (more basic).

c. 0.250 M NH3, Kb = 1.8 × 10-5

Ammonia is a weak base. Therefore, an equilibrium equation and ICE table must be used to determine the pOH, and then the pH of the solution. The Kb value (givn in the problem) is 1.8 × 10-5.

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

INITIAL: The initial concentration of NH3 is given as 0.250 M. The initial concentrations of NH4+ and OH- are zero and the concentration of water is omitted. CHANGE: The change is represented by x and follows the stoichiometry of the balanced chemical reaction. EQUILIBRIUM: The equilibrium row is the sum of the initial and change rows for each column.


NH3 (M) H2O (M) NH4+ (M) OH- (M) I 0.250 M − 0 0 C -x − +x +x E 0.250 - x − x x

The law of mass action is given and the equilibrium values may be substituted into the equation to solve for x and thus the value of [OH-].

> @> @

> @> @> @

> @x0.250xx101.8

NHOHNH

5-

3

-4

b

-

K

u

�

The value of Kb is small, so the assumption that x << 0.250 can be made. > @> @> @0.250

xx101.8 5- u


3

62

101.2x

105.4x�

�

u

u

The value of x is less than 5% of the value from which it was subtracted (0.250) so the assumption is valid. The equilibrium value of [OH-] = x, therefore [OH-] = 2.1×10-3. The pOH of the solution can now be calculated and from that, the pH of the solution since pH + pOH = 14 (at 25°C).

11.32pH68.2 14.00pH

pOH 14.00 pH

2.68pOH)10log(2.1pOH

]log[OHpOH3

-

� �

u�

� �

Given that NH3 is a weak base, a pH value greater than 7 is expected so the answer is reasonable.


Ions as Acids and Bases

1. Question: Label each of the following ions as acidic, basic, or neutral. a. Cl- b. NO2- c. NH4+

d. SO42- e. K+ f. NO3-

Answer:

a. Cl-: The anion is neutral because it does not react with water to any significant amount. Neutral anions can be recognized because there parent acids are strong acids. If Cl- reacted with water, the only product could be HCl which completely ionizes in water, therefore no H+ or OH- ions are generated.

b. NO2-: The anion is basic and it will react with water to produce hydroxide ions. Anions whose parent acid is a weak acid will be basic because they can react with water to produce the acid (which as a weak acid does not completely ionize). The formation of some of the weak acid results in the formation of hydroxide ions.

NO2-(aq) + H2O(l) ⇌ HNO2(aq) + OH-(aq) c. NH4+: The cation is an acid and reacts with water to produce hydronium

ions. The reaction with water will also produce NH3, which is a weak base.

NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) d. SO42-: The anion is basic because it will react with water to produce

HSO42-. The parent acid is HSO4- which is a weak acid. The formation of the weak acid upon reaction of SO42- and H2O also results in the formation of hydroxide ions.

SO42-(aq) + H2O(l) ⇌ HSO4-(aq) + OH-(aq) e. K+: The cation is neutral because it will not react with water to produce

either H+ or OH- ions. f. NO3-: The anion is neutral because will not react with water to produce

either H+ or OH- ions. The parent acid of NO3- is HNO3 which is a strong acid. If the nitrate ion reacted with water, the possible product is a strong acid which completely dissociates in water.


2. Question: Determine whether solutions of the following salts are acidic, basic, or neutral.

a. NaCl b. KNO3 c. KNO2

d. NH4Cl e. NH4F

Answer:

a. NaCl: Neutral; Na+ is a neutral cation and Cl- is a neutral anion so the resulting salt is neutral.

b. KNO3: Neutral; K+ is a neutral cation and NO3- is a neutral anion so the resulting salt is neutral.

c. KNO2: Basic; K+ is a neutral cation and NO2- is a basic anion so the resulting salt is basic.

d. NH4Cl: Acidic; NH4+ is an acidic cation and Cl- is a neutral anion so the resulting salt is acidic.

e. NH4F: Acidic; NH4+ is an acidic cation and F- is a basic anion. The acid-base nature of the salt will be determined by the relative strengths of the acidic cation and basic anion. The Ka of HF is 3.5 × 10-4 and the Kb of NH3 is 1.76 × 10-5. HF is the stronger of the two so it will dominate and the resulting solution is acidic.

3. Question: Calculate the pH of a 0.500 M solution of KNO2. The Ka of HNO2 is 4.3 × 10-4. Answer: The pH of a 0.500 M KNO2 solution can be calculated. Given that it is an ionic compound, two reactions have to be considered; the reaction of K+ or NO2- with water.

K+(aq) + H2O(l) ⇌ No Reaction

If K+ reacts with water, the only possible product is KOH, a strong base that completely dissociates. Therefore, the presence of K+ will not affect the pH of the solution.

NO2-(aq) + H2O(l) ⇌ HNO2(aq) + OH-(aq)


The nitrite ion (NO2-) reacts with water to produce HNO2, a weak acid. Since HNO2 is a weak acid, it does not completely dissociate upon formation which will result in an increase in the [OH-] and therefore change the pH of the solution. This is the only reaction that needs to be considered to determine the pH of the KNO2 solution.

In the reaction, NO2- behaves as a base (proton acceptor) but the the Ka value of HNO2 (4.3 × 10-4) can be obtained from tables. Therefore, Kw = Ka × Kb must be used to determine the value of Kb.

11

414

103.2

103.4100.1�

��

u

uu u

u

b

b

baw

K

K

KKK

Now that the reaction and the equilibrium constant are known, an ICE table can be constructed to find the pH of the solution.

INITIAL: The initial concentration of NO2- is given as 0.500 M. The initial concentrations of HNO2 and OH- are zero and the concentration of water is omitted. CHANGE: The change is represented by x and follows the stoichiometry of the balanced chemical reaction. EQUILIBRIUM: The equilibrium row is the sum of the initial and change rows for each column.

NO2- (M) H2O (M) HNO2 (M) OH- (M) I 0.500 M − 0 0 C -x − +x +x E 0.500 - x − x x

The law of mass action is given and the equilibrium values may be substituted into the equation to solve for x and thus the value of [OH ].


> @> @> @

> @> @> @x0.500

xx102.3

NOOHHNO

11-

2

-2

b

-

K

u

�

The value of Kb is small, so the assumption that x << 0.500 can be made. > @> @> @0.500

xx102.3 11- u


6-

112

104.3x

1015.1x

u

u �

The value of x is less than 5% of the value from which it was subtracted (0.500) so the assumption is valid. The equilibrium value of [OH-] = x, therefore [OH-] = 3.4×10-6. The pOH of the solution can now be calculated and from that, the pH of the solution since pH + pOH = 14 (at 25°C).

8.53pH47.5 14.00pH

pOH 14.00 pH

5.47pOH)10log(3.4pOH

]log[OHpOH6

-

� �

u�

� �

Given that NO2- is acting as a weak base, a pH value greater than 7 is expected so the answer is reasonable.

advanced chemistry practice problems - mrs....

Documents